Subnetting

[rajakajay16]

Hi Guys,
I am practising the subnetting topic and Need a assistance in Usable host part.
Below is the scenario:-
My requirement is for 510 hosts.
IP address:-192.168.1.0/23
My subnet mask is -255.255.254.0
i/p range-192.168.1.1-192.168.1.255
192.168.2.1-192.168.2.254
Here only 509 IP’s are available as per my calculation so I want to know how we calculate the 510 IP as above mentioned range.

[Jon Marshall]

192.168.2.0 is a valid IP with a /23 mask but you may want to avoid using it as it can cause confusion.

Jon

[Richard Burts]

The original post has a misunderstanding that is quite common. It says

i/p range-192.168.1.1-192.168.1.255
192.168.2.1-192.168.2.254

But the range does not start at 192.168.1.1. The range is actually 192.168.0.1 through 192.168.1.254

If we want to get into the details of it then think of the third octet in binary. The mask of 254 is 11111110 which indicates that the first 7 bits of addresses must match and the last bit is variable. Then think of the address in binary which is 00000001. Change the last bit and the other available address is 00000000. So the range begins at 192.168.0 and goes through 192.168.1.

[Jon Marshall]

Rick 

Good point, I was focussed on the fact you can use 192.168.x.0 as an IP but missed that.

Jon

[Joseph W. Doherty]

Actually, to clarify further, using a /23 network, only "odd" numbers for the 3rd octet would be valid host IPs.

e.g.
192.168.0.0/23 invalid host IP
192.168.1.0/23 valid host IP
192.168.2.0/23 invalid host IP
192.168.3.0/23 valid host IP
.
.